Conservation of Linear Momentum

Conservation of Linear Momentum

For a system of particles, define the total linear momentum

(1)

The motion of the entire system is governed by the net external force:

(2)

Integrating over a finite time interval gives

(3)

Therefore, if the net external impulse is zero,

(4)

and the total momentum is conserved:

(5)

This principle is one of the most powerful tools in mechanics because internal forces can be complicated, large, and rapidly varying while the total momentum remains simple.

1 Total momentum of a particle system

For particles indexed by i = 1,…,N,

(6)

The total momentum is

(7)

In Cartesian components,

Px	= ∑ imivix,	(8)
Py	= ∑ imiviy,	(9)
Pz	= ∑ imiviz.	(10)

Momentum conservation is therefore a vector statement. If total momentum is conserved, each Cartesian component is conserved separately:

Px,i	= Px,f,	(11)
Py,i	= Py,f,	(12)
Pz,i	= Pz,f.	(13)

2 Derivation from Newton’s second and third laws

For particle i,

(14)

where F_ij is the internal force on particle i due to particle j.

Sum over all particles:

(15)

The left side is

(16)

For ordinary Newtonian pair forces satisfying Newton’s third law,

(17)

Thus every internal pair cancels from the total force sum:

(18)

Therefore

(19)

3 Internal forces can change individual momenta

Internal-force cancellation does not mean the internal forces vanish.

Two interacting particles can exert large forces on each other:

(20)

Then

	= F12,	(21)
	= F21.	(22)

Their individual momenta change, but

(23)

if no external force acts.

Thus momentum can be transferred internally from one part of a system to another without changing the total.

4 External impulse form

Integrate

(24)

from t_i to t_f:

(25)

Define the external impulse:

(26)

Then

(27)

This is the most general finite-time statement for a fixed set of particles.

5 Conservation condition

If

(28)

then

(29)

A sufficient condition is

(30)

throughout the interval.

However, this is not strictly necessary.

The net external force may be nonzero at some times while the total external impulse still vanishes:

(31)

Thus the impulse condition is more general than the instantaneous-force condition.

6 What is an isolated system?

In elementary mechanics, an isolated system is a system for which external interactions are absent or negligible over the interval being studied.

For momentum analysis, the practically important condition is

(32)

The word “isolated” therefore depends on

• the chosen system boundary,
• the forces acting across that boundary,
• the time interval under consideration,
• the accuracy required.

A system can be approximately isolated for a short collision even if gravity acts continuously.

7 System boundary

Before applying momentum conservation, define what belongs to the system.

Suppose two carts collide.

If the system includes both carts, the contact forces between them are internal and cancel from the total momentum balance.

If the system includes only one cart, the contact force from the other cart is external and changes that cart’s momentum.

Thus the same physical force can be internal or external depending on the chosen system.

A useful rule is:

(33)

8 Center of mass and total momentum

The center-of-mass velocity is

(34)

where

(35)

Therefore

(36)

For a fixed-mass system,

(37)

Hence

(38)

If

(39)

then

(40)

Internal motion can be violent while the center of mass continues uniformly.

9 Example 1: recoil from rest

Two skaters initially stand at rest on nearly frictionless ice.

Let their masses be

m1	= 60 kg,	(41)
m2	= 90 kg.	(42)

They push apart.

Initially,

(43)

After the push,

(44)

Suppose skater 1 moves at

(45)

Then

v2	= −v1	(46)
	= −(3.0)	(47)
	= −2.0 m∕s.	(48)

Thus

(49)

The heavier skater moves more slowly so that the momenta are equal in magnitude and opposite in direction.

10 Recoil does not require equal speeds

For a two-body system initially at rest,

(50)

Therefore

(51)

The momentum magnitudes are equal:

(52)

The speed magnitudes satisfy

(53)

in magnitude.

The lighter object has the larger speed.

11 Example 2: projectile recoil

A launcher of mass

(54)

fires a projectile of mass

(55)

at

(56)

relative to the ground immediately after firing.

Assume the launcher-projectile system is initially at rest and horizontal external impulse is negligible.

Momentum conservation gives

(57)

Thus

V	= −v	(58)
	= −(300)	(59)
	= −3.0 m∕s.	(60)

Therefore

(61)

12 Explosion of a system initially at rest

Suppose an object initially at rest separates into fragments due to an internal energy release.

If external impulse is negligible,

(62)

and therefore

(63)

The fragments can have substantial individual momenta and kinetic energies, but their vector momentum sum remains zero.

The center of mass remains at rest:

(64)

13 Example 3: two-fragment explosion

A body initially at rest explodes into two fragments.

Let

m1	= 2.0 kg,	(65)
m2	= 3.0 kg.	(66)

Fragment 1 moves at

(67)

Momentum conservation gives

(68)

Therefore

v2	= −	(69)
	= −	(70)
	= −4.0 m∕s.	(71)

Thus

(72)

The momentum is conserved even though the final kinetic energy is larger than the initial kinetic energy. The extra kinetic energy came from internal stored energy.

14 Momentum conservation does not imply kinetic-energy conservation

Momentum and kinetic energy are different quantities.

For a two-object isolated system,

(73)

can hold while

(74)

Examples include:

• explosions, where kinetic energy can increase,
• inelastic collisions, where kinetic energy can decrease,
• elastic collisions, where kinetic energy remains constant.

Thus

(75)

15 Componentwise conservation

Because momentum is a vector, conservation applies independently along each coordinate direction.

In two dimensions,

∑ px,i = ∑ px,f,		(76)
∑ py,i = ∑ py,f.		(77)

This is especially useful when external impulse is negligible in one direction but not another.

For example, a puck may experience a large vertical Normal force from a table while horizontal external impulse remains negligible. Horizontal momentum can then be conserved even though the full three-dimensional momentum of the puck-Earth system requires a broader analysis.

16 Example 4: two-dimensional recoil

A system initially at rest separates into three fragments.

Two fragment momenta are

p1	= (4ex) kg m∕s,	(78)
p2	= (3ey) kg m∕s.	(79)

Since

(80)

the third momentum is

p3	= −(p1 + p2)	(81)
	= −4ex − 3ey.	(82)

Thus

(83)

Its magnitude is

(84)

17 Short collisions and negligible external impulse

During a collision, internal contact forces can be extremely large but act for a very short time.

Suppose the collision lasts

(85)

A slowly varying external force such as gravity contributes an impulse of order

(86)

If

(87)

associated with the collision, then gravitational impulse can be neglected during the collision interval.

Thus the system momentum is approximately conserved during impact:

(88)

This approximation is often the reason momentum conservation works so well for short collisions on Earth.

18 Approximate conservation in one direction

Sometimes an external force is large but its impulse is mainly along one direction.

For example, two carts collide on a horizontal track.

The track exerts vertical normal forces, so vertical external forces are not negligible.

However, if horizontal Friction is small,

(89)

Therefore

(90)

Momentum conservation need not be all-or-nothing. It can apply to selected components.

19 Example 5: person throwing an object from a cart

A person and cart have combined mass

(91)

They are initially at rest on a frictionless horizontal track.

The person throws a

(92)

object to the right at

(93)

relative to the ground after release.

Let V be the recoil velocity of the person-cart system.

Initial momentum is zero:

(94)

Final momentum is

(95)

Thus

(96)

Therefore

(97)

The person and cart move left.

20 Relative velocity caution

Momentum conservation uses velocities measured in one common inertial frame.

If a problem gives a projectile velocity relative to a launcher,

(98)

that relative velocity cannot be inserted directly into

(99)

unless the other velocity is expressed in the same frame.

For one-dimensional Galilean motion,

(100)

Always convert to a common inertial frame before summing momenta.

21 Momentum conservation in different inertial frames

Suppose total momentum is conserved in inertial frame S:

(101)

Let frame S′ move at constant velocity V relative to S.

For a fixed-mass system,

(102)

Therefore

Pf′	= Pf − MV,	(103)
Pi′	= Pi − MV.	(104)

Since

(105)

it follows that

(106)

Thus momentum conservation is valid in every inertial frame, even though the numerical value of total momentum changes between frames.

22 Closed versus isolated systems

The words closed and isolated are sometimes used differently across textbooks.

For this mechanics sequence, the safest statement is the equation itself:

(107)

Then:

• if J_ext = 0, total momentum is conserved;
• if J_ext≠0, total momentum changes by exactly that external impulse.

The equations are less ambiguous than terminology.

23 Variable-mass caution

For a fixed set of particles,

(108)

is straightforward.

For an open system where mass crosses the system boundary, momentum is carried with that mass.

A rocket is therefore not analyzed by simply declaring the rocket body alone to have conserved momentum.

One must include momentum flux across the chosen boundary or enlarge the system to include the expelled material.

Variable-mass momentum balance will be developed separately.

24 Common mistakes

1. Conserving the momentum of one object instead of the total system.
2. Forgetting to define the system boundary.
3. Treating internal forces as external forces after both interacting bodies are already included in the system.
4. Assuming zero external force is the only way momentum can be conserved over an interval.
5. Forgetting that zero external impulse, not zero external force at every instant, is the fundamental finite-time condition.
6. Conserving momentum as a scalar when the problem is two- or three-dimensional.
7. Assuming individual momentum components are conserved when external impulse acts along that component.
8. Assuming momentum conservation implies kinetic-energy conservation.
9. Using relative velocities and ground-frame velocities in the same momentum sum.
10. Assuming equal-and-opposite momenta imply equal speeds.
11. Neglecting external impulse during a collision without checking its scale.
12. Applying fixed-mass system equations directly to an open variable-mass system.

25 Practice exercises

1. Two particles have momenta 3e_x and −5e_x in kg m∕s. Find the total momentum.
2. Derive F_ext,net = dP∕dt from Newton’s second and third laws for a two-particle system.
3. Two skaters of masses 50 kg and 75 kg push apart from rest. If the lighter skater moves at 4.0 m∕s, find the heavier skater’s velocity.
4. A 6.0 kg launcher fires a 0.030 kg projectile at 400 m∕s relative to the ground. Find the launcher’s recoil velocity.
5. An object initially at rest explodes into two fragments of masses 1.0 and 4.0 kg. If the lighter fragment moves at 20 m∕s, find the heavier fragment velocity.
6. A three-fragment explosion produces momenta (6, 0) and (0,−8) in kg m∕s for two fragments. Find the third momentum vector and magnitude.
7. Prove that P = MV_CM.
8. Show that if J_ext = 0, then V_CM is constant for a fixed-mass system.
9. Explain why strong internal collision forces do not prevent total momentum conservation.
10. A collision lasts 2.0 ms for a 10 kg two-body system. Estimate the gravitational impulse magnitude and compare it with a collision momentum change of 40 kg m∕s.
11. Give an example in which total momentum is conserved while kinetic energy decreases.
12. Give an example in which total momentum is conserved while kinetic energy increases.
13. Explain why horizontal momentum can be conserved for carts on a track even though vertical external forces are large.
14. A person on a frictionless cart throws an object. Explain why the cart-person system alone does not conserve momentum after the object leaves if the object is excluded from the system.
15. Show that if momentum is conserved in one inertial frame, it is conserved in every Galilean inertial frame for a fixed-mass system.

26 Summary

The total momentum of a particle system is

(109)

Internal Newton’s-third-law force pairs cancel from the total momentum balance, giving

(110)

Integrating,

(111)

If

(112)

then

(113)

The total momentum is related to center-of-mass motion by

(114)

Momentum conservation is vectorial and can be applied separately to individual coordinate directions.

It does not imply conservation of kinetic energy.

These results provide the foundation for recoil, explosions, and collisions. M04-03 applies them directly to one-dimensional collision problems.

References

References